Photoelectric Surface: Work Function & Kinetic Energy

AI Thread Summary
The photoelectric surface has a threshold frequency of 4.62 X 10^14 Hz, leading to a calculated work function of 3.06 X 10^-19 J. For part (b), to find the kinetic energy of the ejected electrons when illuminated with light at a frequency of 5.2 X 10^14 Hz, the work function must be subtracted from the energy of the incoming photons. The energy of the photons can be calculated using the equation E = hf, where h is Planck's constant. The discussion emphasizes understanding the work function's role in determining the kinetic energy of ejected electrons. The solution requires applying the photoelectric effect principles effectively.
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Homework Statement


A certain photoelectric surface has a threshold frequency of 4.62 X 10^14 Hz.
a) What is the photoelectric work function in joules?
b) The surface is illuminated with light that has a frequency of 5.2 X 10^14 Hz. What is the kinetic energy of the electrons ejected from the surface? Answer in joules and electron volts.

Homework Equations


1 eV = 1.6 X 10^19 J
W = hf


The Attempt at a Solution



a) 3.06 X 10^-19 J
b) I need help on! I have no idea which equation to use, etc. Thanks!
 
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potamus said:

Homework Statement


A certain photoelectric surface has a threshold frequency of 4.62 X 10^14 Hz.
a) What is the photoelectric work function in joules?
b) The surface is illuminated with light that has a frequency of 5.2 X 10^14 Hz. What is the kinetic energy of the electrons ejected from the surface? Answer in joules and electron volts.

Homework Equations


1 eV = 1.6 X 10^19 J
W = hf


The Attempt at a Solution



a) 3.06 X 10^-19 J
b) I need help on! I have no idea which equation to use, etc. Thanks!
Welcome to Physics Forums,

You're answer to part (a) is spot on. Now for part (b), what is the definition of the work function?
 

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